Page No 346: - Chapter 9 Ray Optics & Optical Instruments Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 9.5:

A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

Answer:

Actual depth of the bulb in water, d₁ = 80 cm = 0.8 m

Refractive index of water, 

The given situation is shown in the following figure:

Where,

i = Angle of incidence

r = Angle of refraction = 90°

Since the bulb is a point source, the emergent light can be considered as a circle of radius, 

Using Snell’ law, we can write the relation for the refractive index of water as:

Using the given figure, we have the relation:

∴R = tan 48.75° × 0.8 = 0.91 m

∴Area of the surface of water = πR² = π (0.91)² = 2.61 m²

Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m².

Question 9.6:

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Answer:

Angle of minimum deviation,   = 40°

Angle of the prism, A = 60°

Refractive index of water, µ = 1.33

Refractive index of the material of the prism = 

The angle of deviation is related to refractive index as:

Hence, the refractive index of the material of the prism is 1.532.

Since the prism is placed in water, let  be the new angle of minimum deviation for the same prism.

The refractive index of glass with respect to water is given by the relation:

Hence, the new minimum angle of deviation is 10.32°.

Question 9.7:

Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?

Answer:

Refractive index of glass,

Focal length of the double-convex lens, f = 20 cm

Radius of curvature of one face of the lens = R₁

Radius of curvature of the other face of the lens = R₂

Radius of curvature of the double-convex lens = R

The value of R can be calculated as:

Hence, the radius of curvature of the double-convex lens is 22 cm.

Question 9.8:

A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?

Answer:

In the given situation, the object is virtual and the image formed is real.

Object distance, u = +12 cm

(a) Focal length of the convex lens, f = 20 cm

Image distance = v

According to the lens formula, we have the relation:

Hence, the image is formed 7.5 cm away from the lens, toward its right.

(b) Focal length of the concave lens, f = −16 cm

Image distance = v

According to the lens formula, we have the relation:

Hence, the image is formed 48 cm away from the lens, toward its right.

Question 9.9:

An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

Answer:

Size of the object, h₁ = 3 cm

Object distance, u = −14 cm

Focal length of the concave lens, f = −21 cm

Image distance = v

According to the lens formula, we have the relation:

Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.

The magnification of the image is given as:

Hence, the height of the image is 1.8 cm.

If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.

Question 9.10:

What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

Answer:

Focal length of the convex lens, f₁ = 30 cm

Focal length of the concave lens, f₂ = −20 cm

Focal length of the system of lenses = f

The equivalent focal length of a system of two lenses in contact is given as:

Hence, the focal length of the combination of lenses is 60 cm. The negative sign indicates that the system of lenses acts as a diverging lens.

Question 9.11:

A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Answer:

Focal length of the objective lens, f₁ = 2.0 cm

Focal length of the eyepiece, f₂ = 6.25 cm

Distance between the objective lens and the eyepiece, d = 15 cm

(a) Least distance of distinct vision, 

∴Image distance for the eyepiece, v₂ = −25 cm

Object distance for the eyepiece = u₂

According to the lens formula, we have the relation:

Image distance for the objective lens,

Object distance for the objective lens = u₁

According to the lens formula, we have the relation:

Magnitude of the object distance,   = 2.5 cm

The magnifying power of a compound microscope is given by the relation:

Hence, the magnifying power of the microscope is 20.

(b) The final image is formed at infinity.

∴Image distance for the eyepiece,

Object distance for the eyepiece = u₂

According to the lens formula, we have the relation:

Image distance for the objective lens,

Object distance for the objective lens = u₁

According to the lens formula, we have the relation:

Magnitude of the object distance,   = 2.59 cm

The magnifying power of a compound microscope is given by the relation:

Hence, the magnifying power of the microscope is 13.51.

Question 9.12:

A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope,

Answer:

Focal length of the objective lens, f_o = 8 mm = 0.8 cm

Focal length of the eyepiece, f_e = 2.5 cm

Object distance for the objective lens, u_o = −9.0 mm = −0.9 cm

Least distance of distant vision, d = 25 cm

Image distance for the eyepiece, v_e = −d = −25 cm

Object distance for the eyepiece = 

Using the lens formula, we can obtain the value of as:

We can also obtain the value of the image distance for the objective lens  using the lens formula.

The distance between the objective lens and the eyepiece

The magnifying power of the microscope is calculated as:

Hence, the magnifying power of the microscope is 88.

Question 9.13:

A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

Answer:

Focal length of the objective lens, f_o = 144 cm

Focal length of the eyepiece, f_e = 6.0 cm

The magnifying power of the telescope is given as:

The separation between the objective lens and the eyepiece is calculated as:

Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.

Question 9.14:

(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?

(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 10⁶ m, and the radius of lunar orbit is 3.8 × 10⁸ m.

Answer:

Focal length of the objective lens, f_o = 15 m = 15 × 10² cm

Focal length of the eyepiece, f_e = 1.0 cm

(a) The angular magnification of a telescope is given as:

Hence, the angular magnification of the given refracting telescope is 1500.

(b) Diameter of the moon, d = 3.48 × 10⁶ m

Radius of the lunar orbit, r₀ = 3.8 × 10⁸ m

Let  be the diameter of the image of the moon formed by the objective lens.

The angle subtended by the diameter of the moon is equal to the angle subtended by the image.

Hence, the diameter of the moon’s image formed by the objective lens is 13.74 cm

Question 9.15:

Use the mirror equation to deduce that:

(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

(b) a convex mirror always produces a virtual image independent of the location of the object.

(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.

(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

[Note: This exercise helps you deduce algebraically properties of

images that one obtains from explicit ray diagrams.]

Answer:

(a) For a concave mirror, the focal length (f) is negative.

∴f < 0

When the object is placed on the left side of the mirror, the object distance (u) is negative.

∴u < 0

For image distance v, we can write the lens formula as:

The object lies between f and 2f.

Using equation (1), we get:

∴  is negative, i.e., v is negative.

Therefore, the image lies beyond 2f.

(b) For a convex mirror, the focal length (f) is positive.

∴ f > 0

When the object is placed on the left side of the mirror, the object distance (u) is negative.

∴ u < 0

For image distance v, we have the mirror formula:

Thus, the image is formed on the back side of the mirror.

Hence, a convex mirror always produces a virtual image, regardless of the object distance.

(c) For a convex mirror, the focal length (f) is positive.

∴f > 0

When the object is placed on the left side of the mirror, the object distance (u) is negative,

∴u < 0

For image distance v, we have the mirror formula:

Hence, the image formed is diminished and is located between the focus (f) and the pole.

(d) For a concave mirror, the focal length (f) is negative.

∴f < 0

When the object is placed on the left side of the mirror, the object distance (u) is negative.

∴u < 0

It is placed between the focus (f) and the pole.

For image distance v, we have the mirror formula:

The image is formed on the right side of the mirror. Hence, it is a virtual image.

For u < 0 and v > 0, we can write:

Magnification, m  > 1

Hence, the formed image is enlarged.